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Talk:Infinite time Turing machine
I'm pretty sure \(\lambda\) is larger than \(\omega_1^{\text{CK}}\). Deedlit11 (talk) 06:17, October 2, 2013 (UTC) Wait, if \(\Sigma\) is not admissible, is it than equal to or larger than \(\omega_1^{\text{DEF}}\)? Wythagoras (talk) 16:41, October 2, 2013 (UTC) :The first limit of admissible ordinals is \(\omega_\omega^\text{CK}\), so probably \(\Sigma\) is this ordinal. Ikosarakt1 (talk ^ ) 17:02, October 2, 2013 (UTC) ::@Wythagoras: \(\Sigma\) being admissible just means it is not a "round" ordinal. For example, if \(\alpha\) is admissible then \(\alpha + n\) is not admissible for finite \(n\). So no, it doesn't have to be larger than \(\omega_1^{\text{DEF}}\). ::@Ikosarakt1: The fact that \(\lambda\) is recursively inaccessible means it is greater than \(\omega_1^{\text{CK}}\), \(\omega_{\omega_1^{\text{CK}}}^{\text{CK}}\), Goucher's ordinal, and any recursive extension of that. Since \(\Sigma > \lambda\), it is greater than the same ordinals. Deedlit11 (talk) 22:30, October 2, 2013 (UTC) ::Is \(\lambda\) the smallest ordinal which is greater than any recursive extension of \(\alpha \mapsto \omega_\alpha^\text{CK}\)? In that case, lambda is equivalent to \(\omega_{1,2}^\text{CK}\) in my special notation for non-recursive ordinals. Ikosarakt1 (talk ^ ) 23:13, October 2, 2013 (UTC) ::It is very possible that you're right. Such ordinal is called recursively inaccessible, by analogy to inaccessible cardinals. LittlePeng9 (talk) 05:32, October 3, 2013 (UTC) :::It seems that \(\lambda\) is bigger than \(\omega_{1,2}^\text{CK}\). Quoting from Hamkins' and Lewis' paper on Infinite Time Turing Machines, :::Indescribability Theorem 8.3 The supremum λ of the writable ordinals is recursively inaccessible. Indeed, it is the λth recursively inaccessible ordinal, and the λth such fixed point, and so on. This is because λ is indescribable by Π-1-1 properties. Indeed, λ is indescribable by semi-decidable properties. :::The smallest recursively inaccessible ordinal is at least as big as \(\omega_{1,2}^\text{CK}\). (It might be bigger, I'm not sure.) Since λ is the λth recursively inaccessible ordinal, it must be bigger than \(\omega_{1,2}^\text{CK}\). :::The fact that λ is indescribable by decidable (even semi-decidable) properties, might mean it is forever beyond any extension of \(\omega_{1,2}^\text{CK}\) of the form Ikosarakt1 was contemplating. Deedlit11 (talk) 06:16, October 5, 2013 (UTC) Any way we can define fundamental sequences for lambda, zeta, and sigma? Do we observe all ITTMs up to a certain number of states? FB100Z • talk • 19:26, October 2, 2013 (UTC) :Yeah, it looks like setting \(\lambdan =\) the largest writable ordinal using an ITTM with at most \(n\) states is the way to go. (and similarly for \(\zeta\) and \(\Sigma\)). The only concern is whether you actually achieve the ordinal at some finite \(n\). I don't think this is a concern for \(\lambda\) or \(\zeta\) - we might be able to prove that the largest writable/eventually writable ordinal in n states gets larger as n increases - but I'm worried about \(\Sigma\). I think it may be possible that the supremum of accidentally writable ordinals is actually achieved for some ITTM. Deedlit11 (talk) 22:40, October 2, 2013 (UTC) How good are oracle ITTMs? FB100Z • talk • 05:36, October 3, 2013 (UTC) :Very strong. Again quoting from Hamkins' and Lewis' paper, :Jump Closure Theorem 5.4 Every infinite time degree is closed under the Turing jump operator. Indeed, for any real x and any writable ordinal α, the αth Turing jump of x is still infinite time equivalent to x. :So even a single infinite Turing jump will get you farther than any α regular Turing jumps, for any writable ordinal α. Note that there are two natural infinite Turing jumps: Given a real A, we can define :the "strong jump of A" = H^A = {(p,x) | φ^A_p (x) halts} :the "weak jump of A" = A ⊕ h^A = A ⊕ {p | φ^A_p (0) halts} :The strong jump should generate very fast-growing Busy Beaver functions! Deedlit11 (talk) 06:29, October 5, 2013 (UTC) Encoding ordinals as strings On the article I wrote that ordinals are expressible as strings, and here I want to propose such encoding (to verification): Let string 10000... encode 0. Now, if string \(w\) encodes ordinal \(\alpha\), then \(0w\) encodes it's successor (e.g. \(0^n1000...\) encodes finite ordinal n). For limit ordinals, let us have sequence of strings \(w_1,w_2,w_3,...\) encoding fundamental sequence for our ordinal. We use ruler sequence 1213121412131215... to code these strings as a single string: where there are 1's we put string \(w_1\), where there are 2's we put \(w_2\) etc. Now we put a single 1 on the beginning of string, so we won't mess up this with succesor ordinals. I think that using this encoding we can recover original ordinal from any proper encoding. LittlePeng9 (talk) 05:41, October 3, 2013 (UTC) :Interestingly, in the talk of article on \(\omega_1\), Deedlit proved that it is impossible to map all countable ordinals to irrationals in the way preserving order. So, can we disprove it? Ikosarakt1 (talk ^ ) 18:56, October 5, 2013 (UTC) ::Deedlit's proof is valid, as long as by real number ordering we take standard ordering induced by 0<1, a a+c We can create real number ordering such that my ordering, when interpreted as encoding after binary comma, is order preserving mapping f:w_1->(0,1). LittlePeng9 (talk) 20:04, October 5, 2013 (UTC) How large is \(\lambda\)? If the first recursively inaccessible is \(\omega^{CK}_{1,2}\), is the second then \(\omega^{CK}_{1,1,2}\), the third \(\omega^{CK}_{1,1,1,2}\) and the \(\omega\)th \(\omega^{CK}_{122}\)? If it is, \(\lambda\) seems to be equal to any extension using BAN or another array notation of \(\omega^{CK}_{122}\). Wythagoras (talk) 19:31, November 21, 2013 (UTC) If I had to guess, \(\lambda\) is well above that. LittlePeng9 (talk) 19:40, November 21, 2013 (UTC) It seems that \(\lambda\) well exceeds anything based on the function \(\alpha \mapsto \omega_\alpha^\text{CK}\). FB100Z • talk • 20:26, November 21, 2013 (UTC) :Yes, I see. It seems that \(\omega^{CK}_{1\text 3}\) as the \(\omega^{CK}_12\)th recursively inaccessible, and \(\omega^{CK}_{1\text{|}_22}\) is the \(\omega^{CK}_2\)th recursively inaccessible. Wythagoras (talk) 08:18, November 24, 2013 (UTC) (kind of late reply) One of results of Hamkins and Lewis is that \(\lambda\) is not describable by \(\Pi^1_1\) (or even infinite time semidecidable) properties. I'm not 100% sure if that "transadmissible" hierarchy is definable in \(\Pi^1_1\), but it seems quite probable to me. And if that's true - let's say that \(\lambda\) is quite a big ordinal. LittlePeng9 (talk) 15:45, March 5, 2014 (UTC) ITTMs vs Rayo's function How does for example \(\Sigma1000\) compare to the ordinal that describes Rayo's function? Also, can someone find the first ordinal values of the function \(\Sigman\)? Wythagoras (talk) 07:37, February 16, 2014 (UTC) :From what I know, universe of accidentaly writable ordinals is quite uncharged territory, but for me it looks like ITTM's are strong enough to decide FOST formulas. If that's true, I guess Rayo's ordinal would be a writable ordinal. \(\Sigma1000\) looks like a huge ordinal, possibly outgrowing all writable ordinals. I wouldn't go va banque with that, but I say that Rayo's ordinal is small compared to \(\Sigma1000\). LittlePeng9 (talk) 22:19, February 16, 2014 (UTC) ::I would be very surprised if that was true. It seems reasonable that the ITTM ordinals are expressible in first-order set theory... FB100Z • talk • 05:35, February 17, 2014 (UTC) ::LittlePeng9: What if I say \(\lambda1000\)? Do you still think the same? And how about comparing second-order set theory or even type theory to \(\Sigma1000\)? ::FB100Z: What if I say oracle ITTMs? ::Here are my thoughts: ::Let \(\alpha\) be Rayo's ordinal. ::Let \(\beta\) be Fish number 7's ordinal. ::Let \(\gamma\) be the second-order set theory ordinal. ::Let \(\delta\) be the type theory ordinal. ::I would say: \(\lambda1000<\zeta1000<\alpha<\beta<\Sigma1000<\gamma<\delta\) ::I would place oracle ITTMs between \(\gamma\) and \(\delta\). ::Wythagoras (talk) 19:01, February 17, 2014 (UTC) Spot the difference Let \(N(x) = \alpha\) mean "\(x\) is a notation for \(\alpha\)." Consider the three following systems of ordinal notation: * N(an ITTM) = the ordinal written by that ITTM, given blank input (creating a notation up to \(\lambda\)) * N(an ITTM) = the ordinal clocked by that ITTM, given blank input (creating a notation up to \(\gamma\)) * N(an ITTM) = the ordinal eventually written by that ITTM, given blank input (creating a notation up to \(\zeta\)) Unfortunately accidental writing does not produce a unique ordinal. How can we create a notation up to \(\Sigma\)? FB100Z • talk • 05:20, March 6, 2014 (UTC) I'm afraid there is no really satisfying answer. One might consider N(an ITTM) to be the least upper bound of all ordinals accidentally written by that ITTM, but this is a bit meaningless, because then some machines would be a notations for \(\Sigma\), which we probably don't want. There is also a question if every acc. writable ordinal has a notation (it would not be the case if to write some ordinal \(\alpha\) we always had to use an ordinal \(\beta>\alpha\) and then "cut" the notation to leave only some initial fragment), which I don't know answer to. I don't think there are any "obvious" notations for all accidentally writable ordinals. LittlePeng9 (talk) 14:10, March 6, 2014 (UTC) :it's fine if some notations result in \(\Sigma\). we just label them as undefined (as we would do e.g. for non-halting TMs in the \(\lambda\) system). unfortunately your system has a far larger problem: i'm pretty sure it's the same as the one for \(\zeta\) :what if we supply both an ITTM and a time \(\tau\)? we get the notated ordinal by simulating the ITTM and looking at the tape at step \(\tau\). however this probably can't completely notate everything up to \(\Sigma\) due to the circularity of notating \(\tau\) itself FB100Z • talk • 07:30, March 7, 2014 (UTC) : Speaking of that, I realized that the definition of fundamental sequence for \(\Sigma\) is invalid, as there is a single machine (with, say, \(N\) states) which accidentally writes all accidentally writable ordinal, thus making \(\SigmaN=\Sigma\) using our definition, which isn't proper fundamental sequence. Here is such machine: imagine a machine which at once simulates all possible ITTMs (Welsh called it ''universal ''machine), step by step. After each step it reads what real is (accidentally) written on tape of the first machine and copies it to output tape. Then the same with tape of the second machine, and the third etc. and after this we simulate next step of each machine. It's quite obvious that such machine exists and indeed accidentally writes all possible reals. Thus I'm removing the definition of \(\Sigman\). LittlePeng9 (talk) 16:58, March 6, 2014 (UTC) ::Aw shiiiieeeeet. Can we patch this up by disallowing such universal ITTMs? FB100Z • talk • 07:30, March 7, 2014 (UTC)